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The moduli space of unmarked compact genus-$g$ Riemann surfaces. For $g\ge 2$ it is the quotient ${\mathcal{T}}_g/\mathrm{Mod}(S_g)$ of Teichmüller space by the mapping class group, with its quotient topology. The thick subsets are compact by Mumford's compactness theorem.

For covers branched over three real points with a real base point, choose the two generating loops so complex conjugation reverses each. A sheet permutation satisfying the displayed identities lifts that conjugation to an anticonformal equivalence of the covers, including the filled branch points. For closed hyperbolic Riemann surfaces this is a Riemannian isometry. Such an isometry need not come from conjugating the subgroups inside the original deck group, so a Gassmann equivalent pair can be isometric despite nonconjugacy there.

Let $m=\mathrm{ord}(A)$, $n=\mathrm{ord}(B)$ and $r=\mathrm{ord}(AB)$. Cut the Riemann sphere along arcs joining three marked points, say $0,1,\infty$, so the complement is simply connected. Take copies labeled by the elements of $T$ and glue the slit banks with monodromies given by left multiplication by $A,B,(AB{)}^{-1}$. The product relation ensures consistency. The surface is connected because $A,B$ generate $T$. At a puncture with monodromy order $m$, fill each cycle using the local complex coordinate $z=w^m$; use the analogous charts at the other two punctures. This constructs the three-branch-point regular surface cover $N\to {\mathbb{P}}^1$.

Right multiplication by inverses defines a left action of $T$, commutes with the monodromy gluing and extends holomorphically over the filled points. Average a compatible metric over this finite group to obtain a metric for which $T$ acts by Riemannian isometries. When the genus is at least two, the unique compatible curvature-minus-one metric from the uniformization theorem is an invariant choice. Away from the three marked fibers the action is free. The nontrivial stabilizer subgroups are the conjugates of $⟨A⟩$, $⟨B⟩$ and $⟨AB⟩$, of orders $m,n,r$, omitting order-one groups.

There are $|T|/m$, $|T|/n$, $|T|/r$ points over the corresponding branch points. The Riemann-Hurwitz formula givesThis construction also covers the spherical and toroidal cases. If one monodromy has order one it is simply an unbranched marked fiber, with no nontrivial stabilizer.

For the supplied coset actions, faithfulness gives $m=n=3$. Compose permutations with the rightmost factor acting first. The two products $AB$ have cycle decompositionsThus $r=6$. The index is twelve and each subgroup has order eight, so $|T|=96$. ConsequentlyAll cycles of $A,B,AB$ on each coset set have the full orders $3,3,6$. No nontrivial power of any of these branch monodromies fixes a coset. The fixed-coset criterion means that each $U_i$ has trivial intersection with every conjugate of each cyclic branch stabilizer. Since these are all stabilizers on $N$, both $U_i$ act freely. Their quotient surfaces are therefore smooth closed surfaces, not cone-point orbifolds. Sunada theorem supplies isospectrality, and the unbranched Euler characteristic formula givesChoose the hyperbolic metric for these surfaces and their common genus-nine cover.

In this natural construction the two genus-two surfaces are isometric, by an orientation-reversing map. Define the sheet permutationDirect composition verifiesComplex conjugation of the base sphere fixes its three real branch points and reverses the two chosen generating loops, so their monodromies become the inverses. These identities supply a lift, after relabeling sheets by $P$, to an anticonformal map between the two branched covers. It extends across the branch points by the local power charts. An anticonformal map between closed hyperbolic Riemann surfaces is a Riemannian isometry, by uniqueness of the compatible hyperbolic metric.

Equivalently, take the hyperbolic triangle group of signature $(3,3,6)$. Reflection in the side joining the two order-three vertices conjugates its two vertex rotations to their inverses. The displayed sheet permutation identifies the reflected subgroup for the first covering with the subgroup for the second, up to an irrelevant change of base sheet. This is the orientation-reversing equivalence of branched-cover monodromy. The exhibited isometry is anticonformal, so the conclusion is about Riemannian isometry, without asserting that this map is holomorphic. Merely knowing the subgroups were almost conjugate would not have settled isometry; the explicit inverse-monodromy symmetry does. Nor does the data force this symmetry for every arbitrarily chosen invariant deformation of the cover's metric: it is present for the constructed hyperbolic metric.

The effective group is torsion free and every modular cusp is regular in the preceding sense, so orders of a meromorphic weight-$k$ form are integers. At interior points use a local automorphy trivialization; at a modular cusp use the Fourier order of the appropriate slash transform in its cusp width coordinate. Let $C$ be the sum of all modular cusp points, each once, and setA meromorphic function $\phi$ belongs to the Riemann-Roch space $\mathcal{L}(D(f))$ exactly when $\mathrm{div}\phi +D(f)\ge 0$. In the interior this requires $f\phi$ to have no pole; at a modular cusp it requires order at least one. Conversely, the quotient of any weight-$k$ cusp form by $f$ is a meromorphic weight-zero function satisfying precisely those inequalities. This proves the cusp-form divisor presentationTo compute the degree without imposing a valence formula as an extra assumption, use the meromorphic tensor differential $f^{12}(d\tau {)}^{6k}$. Its automorphy factors cancel. Its order at an interior point is $12{\mathrm{ord}}_{P}f$; at a modular cusp it is $12{\mathrm{ord}}_{P}f-6k$, since $d\tau$ is a nonzero constant times $d{q}_c/q_c$. A meromorphic section of the $6k$th tensor power of the canonical bundle has total divisor degree $6k(2g-2)$. The regular-cusp valence formula on a torsion-free modular curve is thereforeFor $k\ge 3$, $\mathrm{deg}D(f)-(2g-2)=(k-2)d/12>0$. The Riemann-Roch theorem says $\ell (D)-\ell (K-D)=\mathrm{deg}D+1-g$, and a divisor of negative degree has no nonzero sections. Thus $\ell (K-D)=0$ andUsing $d=(p^2-1)/2$ and $r_{\infty }=p-1$ givesThe canonical-degree and Riemann-Roch facts used here are general results for compact Riemann surfaces, as permitted.

At a noncritical point a holomorphic map has a local holomorphic inverse. Fibres of a nonconstant map between compact connected Riemann surfaces are finite: zeros are isolated and compactness precludes infinitely many. Around a nonbranch value $w$, choose disjoint inverse neighbourhoods of every point in its fibre. Compactness of the remaining source ensures a sufficiently small disc about $w$ has no additional preimages. Its preimage is therefore a disjoint union of open sets each mapped biholomorphically onto that disc. This proves a finite-sheeted unbranched covering map after removing the branch fibres.

Here the word “regular” must mean locally unbranched or evenly covered, rather than the stronger modern normal-cover condition. A general holomorphic map does not give a normal covering, and the example below itself shows the distinction.

Lift a loop based at $w$ uniquely from each point of $f^{-1}(w)$. Its endpoint defines a permutation; reversing the loop gives the inverse permutation, and concatenation gives composition. Homotopic loops yield the same endpoints. The monodromy group of a covering consists of these permutations. Removing finitely many points from a connected surface leaves it path-connected. Any two points in the same fibre can therefore be joined upstairs; the projection of that path is a loop whose lift joins them. Thus the monodromy action is transitive.

For the rational map of degree three, differentiation givesAt $z=0$, the local degree is three, so a small loop around its value $0$ acts as a 3-cycle. At $z=\pm \sqrt{3}$, the local degree is two; their distinct branch values are $\mp 3\sqrt{3}/2$, and a loop around either acts as a transposition. The poles $\pm 1$ and the point at infinity are simple and unramified. A 3-cycle and a transposition generate the full symmetric group, soThe corresponding degree-three connected cover is not normal: a normal cover would have a regular monodromy action, whereas $S_3$ on three letters has nontrivial point stabilizers. Thus the stronger interpretation of the printed terminology cannot hold.

Consider the singularity of the injective entire function $f$ at infinity. If removable, it is bounded outside a disk and therefore bounded globally; Liouville theorem makes it constant, a contradiction. If essential, the Casorati-Weierstrass theorem says its values outside every large disk are dense. Choose a small neighbourhood $U$ of a finite point; the open mapping theorem puts a nonempty open disk inside $f(U)$. A point outside a disk containing $U$ would map into that open disk, duplicating a value already attained in $U$. This contradicts injectivity. Thus infinity is a pole, so $f$ is a polynomial. If its degree were greater than one, its derivative would have a root; the local power-series form at that point has local degree at least two and cannot be injective. HenceFor a nonconstant holomorphic map of connected compact Riemann surfaces of degree $d$, the Riemann-Hurwitz formula iswhere $g_R,g_S$ are genera and $e_p$ is the local mapping degree, the multiplicity in the target fibre. There are finitely many points with $e_p>1$.

For a degree-two sphere map, the formula gives total ramification two. Since $e_p\le 2$, there are exactly two distinct ramification points, each of local degree two. Their images are distinct, since a fibre has total multiplicity two. Choose a Möbius transformation $T$ taking $0,\infty$ to these domain points and $S_0$ taking their images to $0,\infty$. The resulting rational map has its only zero at zero of order two and its only pole at infinity of order two, hence is $c{z}^2$ with $c\ne 0$. Replace $S_0$ by $S(z)=c^{-1}{S}_0(z)$, obtaining

For a nonconstant holomorphic map $f:X\to Y$ between compact connected Riemann surfaces, its degree iswhere $e_x$ is the ramification index of a holomorphic map at $x$. The valency theorem makes this sum independent of $y\in Y$. The Riemann-Hurwitz formula is

Write $E=\mathbb{C}/\Lambda$. The map $\psi (z+\Lambda )=-z+\Lambda$ is well-defined and biholomorphic, with ${\psi }^2=\mathrm{id}$. Its fixed points satisfyThey are therefore indexed by $\frac{1}{2}\Lambda /\Lambda \cong (\mathbb{Z}/2\mathbb{Z}{)}^2$, so there are four, as in the negation map on a one-dimensional complex torus.

For each $x\in E^{\prime }$ choose a small open disc $U_x$ for which $U_x\cap \psi (U_x)=\varnothing$. The projection restricts to a homeomorphismUse $(p{|}_{U_x}{)}^{-1}$ followed by a holomorphic coordinate on $U_x$ as a chart on $p(U_x)$. Choosing $\psi (U_x)$ instead changes this coordinate by the holomorphic map induced by $\psi$, so the transition functions are holomorphic. These charts make $S^{\prime }$ a Riemann surface and $p:E^{\prime }\to S^{\prime }$ a holomorphic double covering map.

At each omitted fixed point, a centred local coordinate changes by $z\mapsto -z$, so the invariant quotient coordinate is $w=z^2$. Hence the extension has degree two and exactly four ramification points, each with $e_x=2$. Since the complex torus $E$ has genus one, Riemann--Hurwitz givesThus $\boxed{g_S=0}$, in agreement with the general quotient of a one-dimensional complex torus by negation.

The uniformization theorem says that every simply connected Riemann surface is conformally equivalent to exactly one of the Riemann sphere, the complex plane, and the unit disc. A proper simply connected plane domain is noncompact, so it is not the sphere. It cannot be uniformized by the plane: after choosing a point outside the domain, a branch of a square root and then a Möbius transformation would produce a nonconstant bounded entire function, contradicting the Liouville theorem. It is therefore conformally equivalent to the disc, which is the Riemann mapping theorem.

Choose a local coordinate $w$ with $w(p)=0$. Differentiation at $p$ gives a homomorphismIt is injective. Indeed, a nonidentity Möbius transformation fixing $p$ and having derivative one there is parabolic, hence conjugate to a nonzero translation and of infinite order; no such element can belong to the finite group $H$. Every finite subgroup of ${\mathbb{C}}^{\times }$ is a cyclic group of roots of unity, so $H$ is cyclic. Write $n=|H|$.

The averaged coordinatehas derivative one at $p$, so it is a coordinate on a smaller neighbourhood. For $g\in H$, reindexing by $k=hg$ givesAfter shrinking to an $H$-invariant disc $U$, the group therefore acts as all rotations $z\mapsto {\zeta }^{j}z$, where $\zeta$ is a primitive $n$th root of unity. The invariant coordinateidentifies $H\backslash U$ with a disc and gives it a Riemann-surface chart. In these coordinates the quotient map is exactlyThis is the local cyclic quotient of a Riemann surface.

An irreducible complex analytic hypersurface in a complex manifold $X$ is a closed irreducible analytic subset of pure complex codimension one. A local defining function of a complex analytic hypersurface $Y$ at $x$ is a holomorphic function $f$ on a neighbourhood $U$ such thatThe necessary local algebra is that the stalk ${\mathcal{O}}_{X,x}$ is a regular local ring, hence a unique factorization domain, and that the local branches of a hypersurface germ determine finitely many height-one prime ideals. Each is principal; the product of their generators gives $f$, and removing repeated factors makes it reduced. This also covers a globally irreducible hypersurface that has several local branches at a singular point.

A divisor on a complex manifold is a locally finite formal sum $D={\sum }_Y{n}_{Y}Y$ of irreducible analytic hypersurfaces with integer coefficients. On a sufficiently small $U_{\alpha }$, local defining functions give a meromorphic equation $f_{\alpha }$ for $D$. The quotients $f_{\alpha }/f_{\beta }$ are nowhere-zero holomorphic functions. Gluing frames byproduces the holomorphic line bundle associated to a divisor $[D]$, and $f_{\alpha }{e}_{\alpha }$ gives its canonical meromorphic section with divisor $D$.

The Euler sequence on complex projective spaceimplies $\det T{\mathbb{CP}}^n\cong \mathcal{O}(n+1)$. Taking the dual determinant yields the canonical bundle of complex projective spacewhere $H$ is a hyperplane divisor.

The hypotheses on the homogeneous polynomial $p$ say thatis a smooth projective hypersurface of degree $k$, so its divisor line bundle is $[V]\cong \mathcal{O}(kH)$. The Adjunction formula givesThis is the canonical bundle of a smooth projective hypersurface.

Now fix an isomorphism $\Phi :[P]\to [Q]$ and regard $s_P$ and ${\Phi }^{-1}{s}_Q$ as holomorphic sections of the same holomorphic line bundle. They have no common zero because $P\ne Q$. Their homogeneous coordinates therefore define a well-defined holomorphic mapIn a local frame, the quotientis a meromorphic function with divisor $(f)=P-Q$. Thus $F^{-1}(0)=P$ and $F^{-1}(\infty )=Q$, both with multiplicity one. The degree of a holomorphic map $F$ is therefore one. A nonconstant degree-one holomorphic map between compact connected Riemann surfaces is a biholomorphism, so the displayed map is biholomorphic.

If $F:X\to Y$ is a nonconstant holomorphic map between compact connected Riemann surfaces, its local degree at $p$ is the integer $m_F(p)\ge 1$ for which suitable local coordinates giveThe valency theorem states thatis independent of $y\in Y$. This common value is the degree of a holomorphic map, denoted $\mathrm{deg}F$.

Now let $f$ be a nonconstant rational function of degree $d$. If its distinct finite poles have orders $m_1,\dots ,m_r$, and its pole order at infinity is $m_{\infty }\ge 0$, thenThe derivative $f^{\prime }$ has a pole of order $m_j+1$ at each finite pole. When $m_{\infty }>0$, the expansion $f(z)\sim c{z}^{m_{\infty }}$ shows that $f^{\prime }(z)\sim c{m}_{\infty }{z}^{m_{\infty }-1}$ at infinity. Hence the degree of the derivative of a rational function isIn the first case $r\ge 0$, while in the second $1\le r\le d$; thereforeFor every $d\ge 1$, the lower bound is attained by $f(z)=z^d$, whose derivative has degree $d-1$ (with a constant assigned degree zero). For distinct $a_1,\dots ,a_d$, the functionhas $d$ simple poles, degree $d$, and a derivative with $d$ double poles, so $\mathrm{deg}{f}^{\prime }=2d$. Thus both rational bounds are sharp for every $d$.

Let next $g$ be a nonconstant elliptic function for the period lattice $\Lambda$. Its degree is the total order of its poles in a fundamental parallelogram; by the valency theorem, this is also the degree of the induced map $\mathbb{C}/\Lambda \to \hat{\mathbb{C}}$. If these poles have orders $n_1,\dots ,n_s$, then $d=\mathrm{deg}g={\sum }_j{n}_j$, and $g^{\prime }$ has poles of orders $n_j+1$. The degree of the derivative of an elliptic function is consequentlySince every nonconstant elliptic function has at least one pole and $1\le s\le d$,

Let $d\ge 3$ be odd. The Weierstrass elliptic function supplies the lower-bound exampleIt has one pole modulo $\Lambda$, of order $d$, so its derivative has one pole of order $d+1$. For the upper bound, choose distinct points $a_1,\dots ,a_d$ modulo $\Lambda$ and nonzero constants $c_j$ with ${\sum }_j{c}_j=0$. The quasi-periodicity of the Weierstrass zeta function makeselliptic. It has exactly $d$ simple poles, while $g^{\prime }$ has $d$ double poles. Therefore the two bounds are attained for every required odd degree.